how many 3 digits number are divisible by 7?

• 128
  
  1st 3 digit number divisible by 7 = 105
  last 3 digit number divisible by 7 = 994
  
  105,(105+7),(105+2*7),....994
  This is an AP with
  1st term, a1 = 105
  difference, d = 7
  n th term, an = 994
  a1+(n-1)d = an
  105 + (n-1)*7 = 994
  n-1 = 127
  n = 128
  
• 10 years agoHelpfull: Yes(5) No(0)
• first number divisible by 7=105 and last 3 digit no=994
  no of such numbers=(last -first)/(common difference)+1
  =994-105/7+1
  =128
• 10 years agoHelpfull: Yes(1) No(0)
• max num=999
  min num=100
  total 3 digit numbers (999-100)+1=900
  900/7=128 its the short cut no consider the remender
  
• 10 years agoHelpfull: Yes(1) No(0)
• First 3digit no that divisible by 7 is 105 that is 15*7
  and last 3digit no is 994 that 142*7 so
  142-14=128total no divisible by 7
• 10 years agoHelpfull: Yes(0) No(0)
• 15*7=105 is the first 3digit
  142*7=994 is the last 3 digit no
  so 142-14=128no of 3digit no is divisible by 7
• 10 years agoHelpfull: Yes(0) No(0)
• 
  --> Total 3 digit numbers are 1000-100( i.e 1...99 and 1000) =900
  --> 3 digit numbers that divisible by 7 are 900/7 =128
  
  
• 10 years agoHelpfull: Yes(0) No(0)